\(\begin{align}f(x)'=&\left[\frac{\ln x}{x}\right]'\\=&\frac{\left[\ln x\right]'\cdot x-\ln x\cdot [x]'}{x^2}\\=&\frac{\frac{1}{x}\cdot x-\ln x}{x^2}\\=&\frac{1-\ln x}{x^2}\end{align}\)
\(\begin{align}f(x)''=&\left[\frac{1-\ln x}{x^2}\right]'\\=&\frac{[1-\ln x]'\cdot x^2 - (1-\ln x) \cdot [x^2]'}{(x^2)^2}\\=&\frac{-\frac{1}{x}\cdot x^2- (1-\ln x) \cdot 2x}{x^4}\\=&\frac{-x- (1-\ln x) \cdot 2x}{x^4}\\=&\frac{-x-(2x-\ln x \cdot 2x)}{x^4}\\=&\frac{-x-2x+\ln x\cdot 2x}{x^4}\\=&\frac{-3x+\ln x\cdot 2x}{x^4}\\=&\frac{-3+\ln x\cdot 2}{x^3}\end{align}\)
Student, Punkte: 50
─ anonym4e376 28.11.2019 um 07:02